Problem: Seiji Maki doesn't only like to observe relationships being unfolded, he also likes to observe sequences of numbers, especially permutations. Today, he has his eyes on almost sorted permutations.

A permutation a_1, a_2, ..., a_n of 1, 2, ..., n is said to be almost sorted if the condition a_{i + 1} ≥ a_i - 1 holds for all i between 1 and n - 1 inclusive.

Maki is considering the list of all almost sorted permutations of 1, 2, ..., n, given in lexicographical order, and he wants to find the k-th permutation in this list. Can you help him to find such permutation?

Permutation p is lexicographically smaller than a permutation q if and only if the following holds:

  * in the first position where p and q differ, the permutation p has a smaller element than the corresponding element in q.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases.

Each test case consists of a single line containing two integers n and k (1 ≤ n ≤ 10^5, 1 ≤ k ≤ 10^{18}).

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case, print a single line containing the k-th almost sorted permutation of length n in lexicographical order, or -1 if it doesn't exist.

Example

Input


5
1 1
1 2
3 3
6 5
3 4


Output


1 
-1
2 1 3 
1 2 4 3 5 6 
3 2 1 

Note

For the first and second test, the list of almost sorted permutations with n = 1 is \{[1]\}.

For the third and fifth test, the list of almost sorted permutations with n = 3 is \{[1, 2, 3], [1, 3, 2], [2, 1, 3], [3, 2, 1]\}.

Solution (in CPP):

Solution:
// EXPLOSION!
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
#include<chrono>

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpll;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

#define pb push_back
#define mp make_pair
#define rsz resize
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define f first
#define s second
#define cont continue
#define endl '\n'
//#define ednl '\n'
#define test int testc;cin>>testc;while(testc--)
#define pr(a, b) trav(x,a)cerr << x << b; cerr << endl;
#define message cout << "Hello World" << endl;
const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 }; // for every grid problem!!
const ll linf = 4000000000000000000LL;
const ll inf = 1000000007;//998244353    

void pv(vi a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vll a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vector<vi>a) {
	F0R(i, sz(a)) { cout << i << endl; pv(a[i]); cout << endl; }
}void pv(vector<vll>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); }cout << endl; }void pv(vector<string>a) { trav(x, a)cout << x << endl; cout << endl; }
void setIO(string s) {
	ios_base::sync_with_stdio(0); cin.tie(0);
	if (sz(s))
	{
		freopen((s + ".in").c_str(), "r", stdin);
		if (s != "test1")
			freopen((s + ".out").c_str(), "w", stdout);
	}
}
int check(int n, ll k)// check if 1 << n  <  k
{
	if (n > 60)return 1;
	ll val = 1LL << n;
	if (val < k)return 0;
	return 1;
}
void solve(int n, ll k, int c)
{
	assert(check(n - 1, k));
	int j = 2;
	while (j <= n - c + 1)
	{
		if (check((n - j - c + 1), k))
		{
			R0F(i, j - 1)cout << i + c << " ";
			solve(n, k, c + j - 1);
			return;
		}
		k -= (1LL << n - j - c + 1);
		j++;
	}
	R0F(i, j - 1)cout << i + c << " ";

}
void solve()
{
	int n;
	ll k;
	cin >> n >> k;
	if (!check(n - 1, k))
	{
		cout << -1 << endl;
		return;
	}
	solve(n, k, 1);
	cout << endl;
}

int main() {
	setIO("");
	test
	{
		solve();
	}
}